Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $x = \dfrac{k^2 + 3k}{-k^2 - 10k - 21} \div \dfrac{-k^2 + 5k}{-4k^2 + 28k - 40} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $x = \dfrac{k^2 + 3k}{-k^2 - 10k - 21} \times \dfrac{-4k^2 + 28k - 40}{-k^2 + 5k} $ First factor out any common factors. $x = \dfrac{k(k + 3)}{-(k^2 + 10k + 21)} \times \dfrac{-4(k^2 - 7k + 10)}{-k(k - 5)} $ Then factor the quadratic expressions. $x = \dfrac {k(k + 3)} {-(k + 3)(k + 7)} \times \dfrac {-4(k - 5)(k - 2)} {-k(k - 5)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac {k(k + 3) \times -4(k - 5)(k - 2) } { -(k + 3)(k + 7) \times -k(k - 5)} $ $x = \dfrac {-4k(k - 5)(k - 2)(k + 3)} {k(k + 3)(k + 7)(k - 5)} $ Notice that $(k + 3)$ and $(k - 5)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac {-4k(k - 5)(k - 2)\cancel{(k + 3)}} {k\cancel{(k + 3)}(k + 7)(k - 5)} $ We are dividing by $k + 3$ , so $k + 3 \neq 0$ Therefore, $k \neq -3$ $x = \dfrac {-4k\cancel{(k - 5)}(k - 2)\cancel{(k + 3)}} {k\cancel{(k + 3)}(k + 7)\cancel{(k - 5)}} $ We are dividing by $k - 5$ , so $k - 5 \neq 0$ Therefore, $k \neq 5$ $x = \dfrac {-4k(k - 2)} {k(k + 7)} $ $ x = \dfrac{-4(k - 2)}{k + 7}; k \neq -3; k \neq 5 $